Let us take $(y + b)/a = x^2$ and substitute this expression into the unit circle $x^2 + y^2 = 1$: $y^2 + (y + b)/a = 1$; or $ay^2 + y + (b - a) = 0$; or $y = \frac{-1 \pm \sqrt{1 - 4a(b - a)}}{2a}$ (i). If the parabola is to intersect the circle in four distinct points, then we require two real & distinct values for $y$ AND $b > 1$. Our discriminant in (i) needs to be strictly positive for this to occur, or: $1 - 4ab + 4a^2 > 0$; or $(1 + 4a^2)/4a > b > 1$; or $(1 + 4a^2) > 4a$; or $4a^2 - 4a + 1 = (2a - 1)^2 > 0$; or $a \in (0, 1/2) \cup (1/2, \infty)$ (i). For the second part of this problem, let the four abscissae be $x = t, t - d, t - 2d, t - 3d$ (for $0 < t < 1$, and $d > 0$) be in arithmetic progression. Since the parabola and the circle are each symmetric with respect to the $y$-axis, we obtain $t = -(t - 3d)$ and $t - d = -(t - 2d)$, or $d = 2t/3$ (ii). This gives us the abscissae $x = t, t/3, -t/3, -t$ Taking the two solutions for $y$ in (i) will now give us the system of equations: $\frac{-1 + \sqrt{1 - 4ab + 4a^2}}{2a} = a(\pm t )^2 - b$ (iii); $\frac{-1 - \sqrt{1 - 4ab + 4a^2}}{2a} = a(\pm t/3 )^2 - b$ (iv). Adding (iii) and (iv) next yields: $-1/a = (10/9)at^2 - 2b$; or $2b = (10at^2)/9 + 1/a$; or $b = f(a) = (5/9)at^2 + 1/(2a)$ (v). Taking the derivative of (v), $f'(a) = (5/9)t^2 - 1/(2a^2)$, equal to zero yields the critical points $a = \pm 3/[\sqrt{10}t]$ (of which only the positive root is admissible, and the second derivative $f''(a) = 1/a^3$ is positive at this same root, thus minimizing $b = f(a)$). If we return our attention (i), we next check this positive root against the intervals solved for the parameter $a$: CASE I: $a > 1/2$: $3/[\sqrt{10}t] > 1/2$; or $\frac{3\sqrt{10}}{5} > t$ (recall that $0 < t < 1$); and the interval $(0, 1)$ lies entirely within $(0, 3\sqrt{10)}/5)$...hence our requirement for the parameter $a$ is satisfied. CASE II: $0 < a < 1/2$: $0 < 3/[\sqrt{10}t] < 1/2$; (note: the LHS is satisfied for all $0 < t < 1$) or $t > \frac{3\sqrt{10}}{5}$ and $(0, 1) \cap (3\sqrt{10}/5, \infty) = \varnothing$. Thus, only $a > 1/2$ is the desired range. Finally, we compute the minimum value of $b$: $b = f(3/[\sqrt{10}t]) = (5/9)[3/[\sqrt{10}t]]t^2 + (1/2)(\sqrt{10}t/3)$; or $b = [\sqrt{10}/6 + \sqrt{10}/6]t$; or $b = [\sqrt{10}/3]t$ (vi), and since we require $b > 1$ to attain our four intersection points, this yields $t > 3/\sqrt{10}$ (of which the interval $(3/\sqrt{10}, 1)$ lies entirely within $(0, 1)$). Hence, there exists a value of $b$ such that the four intersection abscissae are in arithmetic progression for all $a > 1/2$. QED